3 years ago

7.

Physics

1 answer:

laiz [17]3 years ago

6 0

7. Matthias Schleiden

8. An Element

9. Compound Microscope

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A tractor ploughing a field accelerates at 2 m/s2

Helen [10]

Answer:

3 m/s

Explanation:

Given:

a = 2 m/s²

Δx = 10 m

v = 7 m/s

Find: v₀

v² = v₀² + 2a (x − x₀)

(7 m/s)² = v₀² + 2 (2 m/s²) (10 m)

v₀ = 3 m/s

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3 years ago

Flexibility is earned through regular _____ and ______

grin007 [14]

Answer:

Explanation:

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2 years ago

The core of a certain reflected reactor consist of a cylinder 10 ft high 10 ft in diameter The measured maximum-to-average flux

Westkost [7]

Answer:

The maximum power density in the reactor is 37.562 KW/L.

Explanation:

Given that,

Height = 10 ft = 3.048 m

Diameter = 10 ft = 3.048 m

Flux = 1.5

Power = 835 MW

We need to calculate the volume of cylinder

Using formula of volume

$V =\pi r^2 h$

Put the value into the formula

$V=\pi\times(1.524)^2\times 3.048$

$V= 22.23\m^3$

$V = 22.23\times10^{3}\ Liter$

We need to calculate the maximum power density in the reactor

Using formula of power density

$P=\dfrac{E}{V}$

Where, P = power density

E = energy

V = volume

Put the value into the formula

$P=\dfrac{835\times10^{6}}{22.23\times10^{3}}$

$P=37561.85 = 37.562\times KW/L$

Hence, The maximum power density in the reactor is 37.562 KW/L.

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3 years ago

Which object will be the easiest for a magnet to pull? a 1-gram piece of paper, a 2-gram eraser, a 3-gram steel paper clip, a 4-

Vlad1618 [11]

Steel paper clip because it can be moved by the magnet and it is lighter than the iron nail

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3 years ago

A classroom has 24 fluorescent bulbs, each of which is 32 W. how much energy does it take to light the room for a minute?(unit=J

cestrela7 [59]

Answer:

Energy= 46.08KJ

Explanation:

Given that the power needed to light each bulb is 32W

We know that Power = $\frac{energy}{time}$

The energy needed to light one bulb= $power*time$

Given time = 1minute = 60 seconds

Energy = $32W*60sec$ =1920J

Therefore energy needed to light one bulb is 1920J

The energy needed to light 24 bulbs = $1920*24$ =46080J=46.08KJ

3 0

3 years ago

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